# Is every finite simple group a quotient of a braid group?

Question: Is every finite simple group a quotient of a braid group?

Context: The braid group on two strands $$B_2$$ is isomorphic to $$\mathbb{Z}$$ and so the infinite family of abelian finite simple groups (cyclic of prime order) are all quotients of $$B_2$$.

The braid group on three strands $$B_3$$ is a universal central extension of $$PSL(2, \mathbb{Z})$$, $$B_3 \cong \mathbb{Z} : PSL(2,\mathbb{Z})$$. So the infinite family of finite simple groups $$PSL(2,p)\cong PSp(2,p)$$ are all quotients of $$B_3$$.

This pattern continues with higher $$B_n,n \geq 3$$. There is a quotient of $$B_n$$ which is a (finite index) subgroup of $$Sp(2m,\mathbb{Z})$$ for some $$m$$ ($$m$$ is the rank of the homology of a certain space, details given in this paper https://link.springer.com/article/10.1007/BF02566275.) And moreover if we take this subgroup mod $$p$$ then we get all of $$Sp(2m,p)$$. Thus we have that $$Sp(2m,p)$$ for all primes $$p$$ are quotients of that $$B_n$$. Assuming that every integer is the $$m$$ corresponding to some $$n$$ then we have that every finite simple group $$PSp(2m,p)$$ is a quotient of some Braid group $$B_n$$.

Since $$PSp(2m,p) \cong PSL(2,p)$$ then $$PSL(2,5)$$ and $$PSL(2,7)$$ are indeed quotients of $$B_3$$ and all cyclic groups are quotients of $$B_2$$. The next group to check would be $$A_6$$.

So one way to answer this question would just be to show that $$A_6$$ is not a quotient of any braid group $$B_n$$.

• Since $B_3$ is a central extension of $\mathrm{PSL}(2,\mathbb Z) \cong C_2 * C_3$, a group with trivial centre is a quotient of $B_3$ iff it is $(2,3)$-generated. This includes the vast majority of finite simple groups (though not $A_6$). See mathoverflow.net/questions/365374/…. Jun 28, 2023 at 10:56
• Also, since the pure braid group $PB_3$ (the kernel of the natural map $B_3 \to S_3$) is isomorphic to $F_2 \times \mathbb Z$, and since every finite simple group is $2$-generated, every finite simple group appears as a quotient of $PB_3$ and therefore as a section of $B_3$. Jun 28, 2023 at 12:34

The alternating group $$A_6$$ is not a quotient of any braid group $$B_n$$. This follows from direct calculation. Since $$B_n = \langle x_1, \dots, x_{n-1} \mid [x_i,x_j] = 1 ~ \text{for} ~ |i-j|>1, x_ix_{i+1}x_i = x_{i+1}x_ix_{i+1} ~\text{for}~ 1 \le i \le n-2\rangle,$$ we can enumerate homomorphisms $$B_n \to A_6$$ by counting $$x_1,\dots,x_{n-1} \in A_6$$ satisfying the braid relations. It turns out that $$|\mathrm{Hom}(B_n, A_6)| = \begin{cases}3960 &:n = 3, \\ 5400 &: n=4, \\ 360 &: n = 5. \end{cases}$$ The image of a homomorphism $$B_n \to A_6$$ ($$n \le 5$$) never has order greater than $$60$$. Now observe that there are exactly $$|G| = 360$$ homomorphisms $$B_n \to A_6$$ factoring as $$B_n \to \mathbb Z \to A_6$$. The calculation for $$n=5$$ demonstrates that every homomorphism $$B_5 \to A_6$$ has this form. Since $$B_{n+1}$$ is generated by two copies of $$B_n$$, it follows by induction that every homomorphism $$B_n \to A_6$$ for $$n \ge 5$$ factors through $$\mathbb Z$$ and hence generates a subgroup of order at most $$5$$.
• The same approach works for $A_7$, but you have to calculate $|\mathrm{Hom}(B_6, A_7)|$. Jun 28, 2023 at 13:12
• Same approach works for $A_8$. On the other hand I believe (but I'm not 100% sure) that $A_n$ is $(2,3)$-generated for all $n > 8$, which implies it is a quotient of $B_3$. Therefore $A_6, A_7, A_8$ are the unique alternating groups which are not quotients of braid groups. Jun 28, 2023 at 14:19